1.
2.
3.
4. 5.
Let the required sum be Rs. P. Then Rs. P = Rs. 100 143 Rs.100 143 4 2 = Rs. 1760. 13 5 31 2 1 4 2 To find rate % prt Since I = ∴ r = 100I 100 pt 2 5 Here P = Rs. 468.75, t = 1 or years. 3 3 I = Rs. (500 − 468.75) = Rs. 31.25 ∴rate p.c. = 100 31.25 100 3125 3 = 4 46875 5 468.745 5 3 Rs. 600 f or 2 years = Rs. 1200 for 1 years And Rs. 150 for 4 years = Rs. 600 for 1 year Int. = Rs. 90. ∴Rate = 90 100 = 5 % 1800 1 To find Time prt Since I = . ∴ t = 100I 100 Pr
Here interest = Rs. 15767.50 − Rs. 8500 = Rs. 7267.50 ∴ t = 7267.50 100 = 19 years. 8500 4.5 Let Principal = P, time = t ears, rate = t P Then P t t 100 9 2
∴t =
100
∴t =
9 Race Formula
.
Rate = time =
6.
10 9
100
1
10
9 3 Let the annual payment be P rupees. The amount of Rs. P in 4 years at 5%
3
1 3
1 3 % 3
The amount of Rs. P in 4 years at 5% = 100 4 5P 100 “ “ “ 3 “ = 115P 100
120P 100
1 ∴ rate = 3 % 3
= 110P 100 “ “ “ 1 “ = 105P 100 These four amounts together with the last annual payment of Rs. P will discharge the debt of Rs. 770 115 P 110 P 105 P ∴ 120 P P = 770 100 100 100 100 “
∴
“
550 P
“
2
“
∴
= 770
770×100
= 140
550
100 Hence annual pay ment = Rs. 140
Theorem: The annual payment that will discharge a debt of Rs. A due in t years at the rate of interest r % per 100 AA annum is rt t A 100t 2 Proof: Let the annual payment be x rupees. 100 t 1 r x The amount of Rs. x in (t − 1) yrs at r % = 100 100 t 2 r The amount of Rs. x in (t − 2) yrs at r % = x 100 ----------------------------------------------------------------------------The amount of Rs. x in 2 yrs at r % = 100 2r x 100 The amount of Rs. x in 1 yrs at r % = 100 1r x 100 These (t −1) amounts together with the last annual pay ment of Rs. X will discharge the debt of Rs. A 100 t 1 r 100 t 2 r x+ x ... + 100 r x + x = A ∴ 100 100 100 Or, x [{100 + (t − 1) r} + {100 + (t − 1) r} + ... {100 + r}] = 100A Or, x 100
r t 1 t 2
100 A
mm 1 2 Using the above theorem: 100 770 Annual payment = 54 5 5 100 2
∴x =
100 A rt 1 t 100t 2
Note: 1 + 2 + 3 + ... + m =
7. 8.
770 100 = Rs. 140 550
848 100 = Rs. 200 43 4 4 100 2 Putting the values in the above formula: A 100 80 = 54 5 5 100 2 80 550 Or, A = = Rs. 440 100 By theorem payment =
9.
10.
11.
12.
13.
14.
Let his deposit = Rs. 100 Interest for first 2 yrs. = Rs. 6 Interest for next 3 yrs. = Rs. 24 Interest for the last year. = Rs. 10 Total interest = Rs. 40 When interest is Rs. 40, deposited amount is Rs. 100 ∴ when interest is Rs. 1520, deposited amount = 100 1520 = Rs. 3800 40 Race formula: 1520 × 100 Principal = Intereset ×100 t1r1 + t 2r2 + t3r3 + ... 2 × 3 + 3 × 8 + 1 × 10 1520 × 100 = = Rs. 3800 40 Let the sum it become Rs. 200. ∴ Interest = 200 − 100 = 100 Then, rate = 100 I = 100 ×100 = 10% Pt 100×10 Race formula: Time × Rate = 100 (Multiple number of principal − 1) Multiple number of principal 1 Or, Rate = 100 × time 100 2 1 Using the above formula: rate = = 10% 10 100 3 1 Rate = = 10% 20 Note: A generalized form can be shown as: If a sum of money becomes ‘x’ times in‘t’ years as SI, the rate of interest rate is given by 100 x 1 % t Using the above formula. 100 Multiple number of principal 1 Time = Rate 100 4 1 = = 60 years. 5 Amount of 1st part = 110 1st part 100 nd “ “ 2 part = 115 2nd part 100 120 rd 3rd part “ “ 3 part = 100 According to t he question, these amounts are equal ∴ 110 × 1st part = 115 × 2nd part = 120 × 3rd part ∴ 1sr part: 2nd part: 3rd par t= 1 : 1 : 1 110 115 120 Hence, dividing Rs. 2379 into three parts in the ratio 276: 264: 253, we have 1st part = Rs. 828, 2nd part= Rs. 792, 3rd part = Rs. 759. P + SI for 3.5 yrs = Rs. 873 P + SI for 2 yrs = Rs. 756 On subtracting, SI for 1.5 yrs = Rs. 117 Therefore, SI f or 2 yrs = Rs. 117 2 = Rs. 156 1.5
15.
16.
17.
18.
∴ P = 756 − 156 = Rs. 600 And rate = 100 156 = 13% per annum 600 2 Let the sum be Rs. x and the original rate be y% per annum. Then, new rate = (y + 3) % per annum. x y 3 2 x y 2 ∴ = 300 100 100 xy + 3x − xy = 15000 or x = 5000 Thus, the sum = Rs. 5000 Race formula: Direct Formula Sum = More Intereset × 100 300 100 = 5000 2 3 Time × More rate Rate =
100(2 1) 100 7 7
Time
100(4 1 21years 100 7
OTHER METHOD: This question can be solved without writing anything. Think like. Doubles in 7 years Trebles in 14 years 4 times in 21 years 5 times in 28 years And so on. Let the amount lent at 3% rate be Rs x, then 3% of x + 5% of (4000 – x) = 144 Or, 3x + 5 4000 – 5x = 14400 Or, 2x = 5600 X = 2800 Thus, the two amounts are Rs 2800 and (4000–2800) or Rs 1200 First rate of interest =
120 100 800 3
= 5%
New rate = 5 + 3 = 8% New interest =
800 3 8 Rs192. 100
New amount = 800 + 192 = 992.
19.
Simple interest for 5 years = Rs 300 Now, when principal is trebled, the simple interest for 5 years will also be treble the simple interest On original principal for the same period. Thus SI for last 5 years when principal is trebled =3 300 = Rs 900 Total SI for 10 yrs = 300 + 900 = Rs 1200
Theorem: A sum of Rs X is lent out in n parts in such a way that the interest on first part at r1% for t1 yrs, the interest on Second part at r2% for t 2 yrs the interest on third part at r3% for t3 yrs. And so on, are equal, the ratio in which the sum was divided in n parts is
given by
1 1 1 1 : : : ............ . r1t1 r2t 2 r3t 3 rn tn
Proof: Let the sum be divided into S1 , S2 ............Sn int 100 r1t1 int 100 S2 r 2t 2 int 100 S3 r 3t 3 int 100 Sn rn tn S1
[Since interest of all parts are equal] S1 : S2 : S3 : ........Sn
= 20.
1 1 1 1 : : : ............ r1t1 r2 t2 r3 t3 rn tn
1stpart 5 10 2ndpart 6 9 100 100 1stpart 6 9 27 27 : 25 2ndpart 5 10 25 2600 1stpart 27 Rs1350 27 25 and 2ndpart 2600 1350 Rs1250
Each Interest =
Or, 21.
int100 int100 int100 int100 : : ................. : r1t1 r 2t 2 r 3t 3 rn tn
Interest = Rs 840 – Rs 750 = 90 Time =
90 100 3 yrs 750 4
Now, by the formula: Sum =
100 amount 100 575 Rs500 100 rt 100 3 5
Note: There is a direct relationship between the principal and the amount And is given by Sum 22.
100 Amount 100 rt
Use the formula Principal = Sum
100 Amount 100 2613 100 2613 Rs 2010 100 rt 100 30 130
Again by using the same formula: 100 3015 100 5t 100 3015 or ,100 5t 2010 1 100 3015 100 2010 t 5 2010 100 (3015 2010) 100 1005 100 1005 10 years 2010 5 2010 5 2010 5 2010
23.
Let the sum be Rs X. x 8 4 32 x 100 100 32 x 68 x x 100 100 68 x interest is less, 100
Interest =
When
the sum is Rs X.
x 100 340 Rs500 68 x 100 100 340 Sum 340 Rs500 100 8 4 68
When interest is 340less, the sum is
Direct formula: 24.
We may consider that Rs (1800 – 1650) gives interest of Rs 30 at 4% per annum. Time
25.
30 100 5 yrs. 150 4
After 2 yrs amount returned to Ramu 400
400 5 2 Rs 440 100
Amount returned to Arun = 2% of Rs 440 = Rs 8.80.
26.
Theorem: When different amounts mature to the same amount at simple rate of interest, the ratio of the amounts Invested are in inverse ratio of (100 + time rate). That is, the ratio in which the amounts are invested is 1 1 1 1 : : : .................. : 100 r1t1 100 r2t2 100 r3t3 100 rntn
Proof: We know that sum =
100 amount 100 rt be S1 , S2 ............Sn ,
Let the sum invested At the rate of r1 , r2 , r3 ..................rn for time t1 , t2 , t3 ,
tn yrs respectively Then
S1 : S2 : S3 :..................: Sn 100 A 100 A 100 A 100 A : : : ....................... : 100 r1t1 100 r2t2 100 r3t3 100 rn tn
[Since the amount (A) is the same for all] 1 1 1 1 : : : ....................... : 100 r1t1 100 r2t2 100 r3t3 100 rn tn
Soln: Therefore, the required ratio is this case is 1 1 1 1 1 1 : : : : 100 2 5 100 3 5 100 4 5 110 115 120
27.
28.
Doubles in 4 yrs 3 times 4 2 = 8 yrs 4 times 4 3 = 12 yrs 8 times 4 7 = 28 yrs Thus direct formula: X Times in = No. of yrs to double (X – 1) 8 times = 4(8 – 1) = 4 7 = 28 yrs. Let the sum be Rs X, then x 15 5 x 15 7 144 100 200 or ,150 x 105 x 144 200 x
144 200 Rs 640 45
Direct formula: Two equal amounts of money are deposited at r1% and r2% for t1 and t2 yrs respectively. If the difference between Their interest is id then sum = Thus, in this case: sum =
id 100 r1t1 r2 t2
144 100 144 100 Rs640 15 5 15 3.5 22.5
500 2 r1 10r1 100 500 2 r2 I2 10r2 100 I1 I 2 10r1 10r2 2.5 I1
29.
Or, r1–r2 =
2.5 10
= 0.25%
By Direct formula: When t1 = t2, (R1–r2) = 30.
I d 100 2.5 100 0.25% Sum t 500 2
Let the sum be Rs X, then at 4% rate for 4 yrs the simple interest =
x 4 4 4x Rs 100 25
At 5% rate for 3 yrs the simple =
x 5 3 3x Rs 100 20
4 x 3x 80 25 20 16 x 15 x Now, we have, or , 80 100 X Rs8000
Quicker Method: For this type of question Sum = 31.
Difference 100 80 100 = Rs 8000 r1t1 r2t2 4 4 3 5
Suppose the of interest = r% and the sum = Rs A A A r 4 600; or, A + r 600 100 25 r or , A 1 600 (1) 25 A r 6 and , A 650; 100 r or , A 1 600 25 A r 6 and , A 650; (2) 100
Now, A +
Dividing (1) by (2), we have r 25 600 ; Or , (25 r ) 2 12 3R 650 50 3r 13 1 50 Or, (50 + 2r) 13 = (50 + 3r) 12 1
Or, 650 + 26r = 600 + 3r; or, 10r = 50 r = 5% Direct formula: If a sum amounts to Rs A1 in t1 years and Rs A2 in t2 years at simple rate of interest, Then rate per annum =
100 A2 A1
A1t2 A2t1
In the above case, R= 32.
100 650 600
6 600 4 650
100 50 5% 1000
Any sum that is paid back to the bank before the last instalment is deducted from the principal and not from the interest. Total interest = Interest on Rs 7000 for 3 yrs + Interest on (Rs 7000 – Rs 3000) = Rs 4000 for 2 yrs. Or, (5450 + 3000 – 7000) =
7000 3 r 4000 2 r 100 100
Or, 1450 = 210r + 80r 33.
r
1450 290
= 5%
Suppose Rs X was lent at 6% per annum. x 6 5 (7000 x) 4 5 1600 100 100 3x 7000 x or , 1600 10 5 3x 14, 000 2 x or , 1600 10
Thus,
X = 16000 – 14000 = Rs 2000.
By Method of Alligation: Overall rate of Interest =
6%
1600 100 32 % 5 7000 7
4% 32 % 7
4 % 7
34. 35.
Ratio of two amount = 2: 5 7000 Amount lent at 6% = 2 =Rs 2000 7 400 5 6 SI = = Rs 120 100
1 146 15 1 4 365 4 100 1225 2 15 1 = Rs. 4 5 4 100 147 = Rs. = Rs. 4.59 nearly. 32
Interest = Rs. 306
10 % 7
